Bài tập 36 trang 163 SGK Toán 11 NC

a)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^3} - 5}}{{{x^2} + 1}} = \mathop {\lim }\limits_{x \to  + \infty } x.\frac{{{x^2}\left( {1 - \frac{5}{{{x^3}}}} \right)}}{{\left( {1 + \frac{1}{{{x^2}}}} \right)}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } x.\frac{{1 - \frac{5}{{{x^3}}}}}{{1 + \frac{1}{{{x^2}}}}} =  + \infty 
\end{array}\)

(vì \(\mathop {\lim }\limits_{x \to  + \infty } x =  + \infty ,\mathop {\lim }\limits_{x \to  + \infty } \frac{{1 - \frac{5}{{{x^3}}}}}{{1 + \frac{1}{{{x^2}}}}} = 1 > 0\))

b) Với mọi x < 0, ta có: 

\(\frac{{\sqrt {{x^4} - x} }}{{1 - 2x}} = \frac{{{x^2}\sqrt {1 - \frac{1}{{{x^3}}}} }}{{1 - 2x}} = \frac{{\sqrt {1 - \frac{1}{{{x^3}}}} }}{{\frac{1}{{{x^2}}} - \frac{2}{x}}}\)

Vì \(\mathop {\lim }\limits_{x \to  - \infty } \sqrt {1 - \frac{1}{{{x^3}}}}  = 1,\mathop {\lim }\limits_{x \to  - \infty } \left( {\frac{1}{{{x^2}}} - \frac{2}{x}} \right) = 0\) và \(\frac{1}{{{x^2}}} - \frac{2}{x} > 0\) với x < 0)

Do đó \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^4} - x} }}{{1 - 2x}} =  + \infty \)

-- Mod Toán 11 HỌC247